# Thread: More questions on parametic curves in 2D

1. ## More questions on parametic curves in 2D

Hi

My question has two parts:
Using the following two equation, $\displaystyle x=lnU$ & $\displaystyle y=ln(\frac{2}{U}$ find:
1) calculate $\displaystyle \frac{d^2y}{dx^2}$

This is what i have done:

find $\displaystyle \frac{dy}{dx}$ first

$\displaystyle \frac{dy}{dx} = \frac{dy}{dU} \frac{1}{\frac{dx}{dU}}$
$\displaystyle \frac{dy}{dx} = \frac{-2U^{-2}}{2U^{-1}} \frac{1}{\frac{1}{U}}$
$\displaystyle \frac{dy}{dx} = \frac{-u}{u}$
$\displaystyle \frac{dy}{dx} = -1$

$\displaystyle \frac{d^2y}{dx^2} = -U$

where have a gone wrong?

2) eliminate the parameter to give a direct relation between x and y:

tried this way but was wrong.
$\displaystyle e^x = U$

$\displaystyle y=ln(\frac{2}{e^x}$

answer in book is $\displaystyle y=ln2-x$
P.S

2. $\displaystyle \displaystyle \log\left( \frac{a}{b}\right)=\log a-\log b$,

any base.

3. Hello, Paymemoney!

Given: .$\displaystyle \begin{Bmatrix} x &=& \ln u \\ y &=& \ln(\frac{2}{u}) \end{bmatrix}$

1) Find: .$\displaystyle \dfrac{d^2y}{dx^2}$

Note that: .$\displaystyle u > 0$

We have: .$\displaystyle y \:=\:\ln2 - \ln u \quad\Rightarrow\quad \dfrac{dy}{du} \:=\:-\dfrac{1}{u}$

And: .$\displaystyle x \:=\:\ln u \quad\Rightarrow\quad \dfrac{dx}{du} \:=\:\dfrac{1}{u}$

Hence: .$\displaystyle \displaystyle \frac{dy}{dx} \;=\;\frac{\frac{dy}{du}}{\frac{dx}{du}} \;=\;\frac{\text{-}\frac{1}{u}}{\frac{1}{u}} \;=\;-1$

Therefore: .$\displaystyle \dfrac{d^2y}{dx^2} \;=\;0$

2) Eliminate the parameter to give a direct relation between $\displaystyle \,x$ and $\displaystyle \,y.$

Answer in book is: $\displaystyle y\:=\:\ln2-x$

We have:. . $\displaystyle \begin{array}{ccccccccc} x &=& \ln u & [1] \\ y &=& \ln(\frac{2}{u}) & \Rightarrow & y &=& \ln 2 - \ln u & [2] \end{array}$

Substitute [1] into [2]: .$\displaystyle y \;=\;\ln 2 - x$